In the given potentiometer circuit arrangement,the balancing length $AC$ is measured to be $250 \, cm$. When the galvanometer connection is shifted from point $(1)$ to point $(2)$ in the given diagram,the balancing length becomes $400 \, cm$. The ratio of the emf of two cells,$\frac{\varepsilon_{1}}{\varepsilon_{2}}$ is -

The balancing length for a cell is $560 \; cm$ in a potentiometer experiment. When an external resistance of $10 \; \Omega$ is connected in parallel to the cell,the balancing length changes by $60 \; cm$. If the internal resistance of the cell is $\frac{N}{10} \; \Omega$,where $N$ is an integer,then the value of $N$ is:

Potentiometer measures the potential difference more accurately than a voltmeter because

With the help of a potentiometer,we can determine the value of the emf of a given cell. The sensitivity of the potentiometer is:
$(A)$ directly proportional to the length of the potentiometer wire
$(B)$ directly proportional to the potential gradient of the wire
$(C)$ inversely proportional to the potential gradient of the wire
$(D)$ inversely proportional to the length of the potentiometer wire
Choose the correct option for the above statements:

The figure shows a $2.0 \; V$ potentiometer used for the determination of internal resistance of a $1.5 \; V$ cell. The balance point of the cell in open circuit is $76.3 \; cm$. When a resistor of $9.5 \; \Omega$ is used in the external circuit of the cell,the balance point shifts to $64.8 \; cm$ length of the potentiometer wire. Determine the internal resistance (in $\Omega$) of the cell.

$A$ uniform conducting wire $AB$ of length $5 \, m$ and resistance $5 \, \Omega$ is connected as shown in the circuit. If the balancing point is obtained at $3 \, m$ from $A$, then the value of $E$ is. (in $ \, V$)